Find the number of real values of $x$ such that $x^2 + 10000[x] = 10000x$ ($[]$ is the floor function) 
Find the number of real values of $x$ such that $x^2 + 10000[x] = 10000x$ ($[]$ is the floor function)

What I Tried: If $x$ is an integer then :-
$$x^2 = 0$$
$$\rightarrow x = 0$$
So $x = 0$ is the only integer solution, the problem with me is to look for real solutions. I have no clear idea on how to start doing it. One thing I did was let $[x] = x - \{x\}$ . We have :-
$$x^2 + 10000(x - \{x\}) = 10000x$$
$$\rightarrow x^2 - 10000\{x\} = 0$$
$$\rightarrow x^2 = 10000\{x\}$$
Can anyone help me here?
 A: Let
$$x = n + r, \; n \in \mathbb{Z}, \; 0 \le r \lt 1 \tag{1}\label{eq1A}$$
Your equation then becomes
$$\begin{equation}\begin{aligned}
x^2 + 10000\lfloor x\rfloor & = 10000x \\
(n + r)^2 + 10000n & = 10000(n + r) \\
n^2 + 2nr + r^2 - 10000r & = 0
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Treat the above equation as a quadratic in $n$ and use the quadratic formula to get
$$\begin{equation}\begin{aligned}
n & = \frac{-2r \pm \sqrt{(2r)^2 - 4(r^2 - 10000r)}}{2} \\
& = \frac{-2r \pm \sqrt{40000r}}{2} \\
& = -r \pm 100\sqrt{r}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Note $-r - 100\sqrt{r}$ is a strictly decreasing function, going from $0$ down to just greater than $-1 - 100 = -101$, i.e., $(-101, 0]$, so the integer values are $-100 \le n \le 0$. With $-r + 100\sqrt{r}$, it's a strictly increasing function, going from $0$ to just less than $-1 + 100 = 99$, i.e., $[0, 99)$, so the integer values are $0 \le n \le 98$. Altogether, the integer values are $-100 \le n \le 98$, with just one value of $r$ for each integer $n$, giving a total of $199$ solutions.
